Continuity

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close as you want

A function being continuous at a domain number is the opposite of being discontinuous.

Function values are close when the domain numbers are close.

Continuity

The function $f$ is continuous at $a$ if

$a \in Dom_f$ - $a$ is in the domain of $f$ , which means $f(a) \in \mathbb {R}$ , i.e. $f(a)$ exists.
For every $\epsilon > 0$ defining an open interval around $f(a)$ , namely $I = (f(a) - \epsilon , f(a) + \epsilon )$ , there exists a corresponding $\delta > 0$ , such that the image of open interval $(a - \delta , a + \delta )$ is a subset of $(f(a)-\epsilon , f(a)+\epsilon )$ .

For EVERY open interval $I$ around $f(a)$ , there exists a corresponding open interval around $a$ , such that ALL of the function values are inside $I$ .

$\blacktriangleright$ Discontinuity: You can find a single interval around $f(a)$ , such that EVERY interval around $a$ contains a domain number whose function value is outside that interval.

$\blacktriangleright$ Continuity: For any and every selected interval around $f(a)$ , you can find a corresponding interval around $a$ , such that all of their function values are inside the selected interval.

$\text {Let } \, f(x) = \begin {cases} 2x-4 & \text { on } (-2, 5) \\ -x + 9 & \text { on } [5, 9) \end {cases}$

Graph of $y=f(x)$

$f$ is continuous at $3$ .

First, $f(3) = 2$ .

Pick ANY open interval around $2$ : $(2 - \epsilon , 2 + \epsilon )$ , where $\epsilon > 0$ is small.

Then there is a corresponding interval for the domain. Namely choose $\delta = \frac {\epsilon }{4}$ .

Let $I = (3 - \delta , 3 + \delta ) = \left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )$

The image of $I$ , $f(I) = f\left (\,\left (3 - \frac {\epsilon }{4}, 3 + \frac {\epsilon }{4}\right )\,\right ) = \left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right )$

And, $\left (2 - \frac {\epsilon }{2}, 2 + \frac {\epsilon }{2}\right ) \subset (2 - \epsilon , 2 + \epsilon )$

ALL of the function values land inside the original $(2 - \epsilon , 2 + \epsilon )$ interval.

This can be done FOR ANY $\epsilon > 0$ .

For ANY open interval around $f(a)$ , no matter how small, a corresponding open interval around $a$ can be found, such that the function value from that interval around $a$ never jump out of the interval around $f(a)$ .

Continuity: No matter what $\epsilon$ is choosen for $f(a)$ , you can ALWAYS find a corresponding $\delta$ for $a$ .

Discontinuity: There is an $\epsilon$ for $f(a)$ , such that there is no $\delta$ for $a$ .

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more examples can be found by following this link
More Examples of Continuity